Complex Analysis Homework 4: Solutions

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1 Complex Analysis Fall 2007 Homework 4: Solutions a The function fz 3z 2 +7z+5 is a polynomial so is analytic everywhere with derivative f z 6z + 7. b The function fz 2z is a composition of polynomials so is analytic everywhere with derivative f z 82z by the chain rule. c The function fz 3z 1/3 z is rational and so is analytic where z 3. By the quotient rule, its derivative is f 8 z 3 z Let fz z and write f u + iv and z x + iy. Then u x 2 + y 2 and v 0. We find that u x x2 + y 2, u y y x2 + y 2 v v y 0. In order for f to be analytic the Cauchy-Riemann equations must hold. That is, we need x x2 + y 2 y x2 + y 2 0 which is impossible, since it requires x and y to be both simultaneously zero and nonzero. Therefore f is differentiable nowhere. Here s another proof. Since the partial derivatives of Re z x 2 + y 2 do not exist at 0, 0, the Cauchy-Riemann theorem implies that z cannot be differentiable at z 0. Suppose that fz z were analytic at some z 0 C, z 0 0. Then fz 2 z 2 zz would be analytic at z 0 and, since z 0 is nonzero, so too would be fz 2 /z z. But we know that gz z is analytic nowhere, so this is impossible. Therefore fz z cannot be analytic at any nonzero complex number either. Hence, z is differentiable nowhere, i.e. is not analytic a According to Cauchy-Riemann equations, if f is analytic then f 1 i y 1

2 so that z i y 2 f + f f. b If fz z x + iy then f is analytic so that by part a we have z f 1. Moreover so that 1, y i z f i y 2 c If fz z x iy then Therefore and 1, y i. z i y 2 z i y 2 d We show first that and satisfy the sum, product and scalar multiple rules when y applied to complex valued functions. Let f, g be complex valued functions and let c be a complex scalar. Write f s + it, g u + iv, c a + ib. Scalar multiplication. We have cf as bt + iat + bs so that by definition and the linearity of derivative of real functions we have cf as bt a s b t a + ib c at + bs + i a t + i s + i t + b s which proves the scalar multiplication rule for. The proof for is identical, replacing y x by y throughout. 2

3 Sum. We have f + g s + u + it + v so that by definition and the linearity of the derivative of real functions we have f + g s + u s + u s + i t + g t + v + i t + i + v u + + iv which proves the addition rule for. The proof for y throughout. is identical, replacing x with y Product. We have fg su tv + itu + sv so that by definition and the product rule for derivatives of real functions we have fg su tv s u + u s tu + sv + i tv v t + i t u + u t + sv + v s s u tv + i t u + sv + u s v t u s s + it + iv + u + iv + i t f g + g which proves the product rule for. The proof for y throughout. + i u t + v s is identical, replacing x with y Finally, we are in a position to verify that the sum, product and scalar multiplication rules hold for and. We prove a slightly more general statement that includes both z z of these statements as special cases. Let a, b C and set D c +b. I claim that D satisfies the sum, product and scalar y multiple rules. Let f, g be complex functions and let c C. By what we have already shown we have Dcf a cf ac + bcf y + bc y cdf 3

4 which proves the scalar multiple rule. Also proving the sum rule. Finally f + g Df + g a a + g f + g + b y + b y + g y a + b y + a g + bg y Df + Dg, Dfg a fg + bfg y a f g + g f a g + bg + g y fdg + gdf, which establishes the product rule for D. Now we are finished since and are both of the form dealt with above. + b z i y z i y f g y + g y a + b y e This is the easy direction. If a nm 0 whenever m 0 we have n0 m0 a nm z n z m a n0 z n z 0 n0 which is a polynomial in z and hence analytic everywhere. a n0 z n For this we use the result of exercise 13, which says that if fz is analytic then z 0. Since z obeys the product rule and z z for m Z +. Moreover, since z n is analytic we know that zn z n0 1 an easy induction shows that zm z mz m 1 0. Therefore, if the 4

5 expression in question is analytic we have by the sum, product and scalar multiple rules 0 N a nm z n z m z n0 m0 n0 m0 n0 m0 n0 m0 z a nmz n z m a nm z n zm zn + zm z z a nm mz n z m 1 z m. But the only way that a polynomial can be identically zero is if all of its coefficients vanish. That is, we need ma nm 0 for all m and n. In particular, if m 0 we must have a nm 0, which is exactly what we sought to prove Before we begin, notice that the statement aux, y + bvx, y c with not all of a, b, c equal to zero is equivalent to saying that the values of fz ux, y + ivx, y lie on a straight line. a We write u ux, y, v vx, y etc. to simplify notation. If we apply / and /y to the equation au + bv c we obtain a u + bv 0 a u y + bv y 0. Since f is analytic we can apply the Cauchy-Riemann equations in the second equality above to obtain the system or, in matrix form, a b b a a u + bv 0 b u av 0 u v 0 0. The determinant of the coefficient matrix is a b + b 2 which cannot be zero because a, b, c are real and not all zero. Hence, the only solution to the system is u v 0. 5

6 Therefore f z u + iv 0 and, since A is connected, we conclude that f is constant. b If a, b, c are complex, then the proof above does not apply. However, if we write a a 1 + ia 2, b b 1 + ib 2 and c c 1 + ic 2 with a 1, a 2, b 1, b 2, c 1, c 2 R, then the equation au + bv c is equivalent to the pair of equations a 1 u + b 1 v c 1 a 2 u + b 2 v c 2. Since not all of a, b, c are zero, it must be that in at least one of these equations not all of the constants are zero. We may then apply part a to that equation to conclude that f is constant. In other words, the statement is valid for complex a, b, c as well Let A C be connected and open and let f : A C be an analytic function. We are asked to prove the following: if fn for some n Z + 0 then f is a polynomial of degree n. We induct on n. When n 0 the statement we need to prove is: if f 0 then f is constant. This is Proposition and was proven in class. We now assume the statement holds for some n 0. Suppose that f n+2 0. Then f n+1 f n+2 0 so that, by our assumption, f is a polynomial of degree n. Write with all a k C. Let Then hz is analytic on A and f z hz fz h z f z a k z k k0 k0 a k k + 1 zk+1. a k z k 0. But we have already seen that this implies h is constant. If h c C then we see that fz hz + k0 k0 a k k + 1 zk+1 c + k0 a k k + 1 zk+1, i.e. f is a polynomial of degree n + 1. Therefore, if the statement fn implies f is a polynomial of degree n holds for any n 0 it also holds for n + 1. By induction, we conclude that the statement is true for all n 0. 6